How do you solve #\frac { 9} { 4} = \frac { 12} { 4x + 1}#?

2 Answers
May 10, 2017

See a solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(4)color(blue)((4x + 1))# to eliminate the fractions while keeping the equation balanced. #color(red)(4)color(blue)((4x + 1))# is the Lowest Common Denominator of the two fractions:

#color(red)(4)color(blue)((4x + 1)) xx 9/4 = color(red)(4)color(blue)((4x + 1)) xx 12/(4x + 1)#

#cancel(color(red)(4))color(blue)((4x + 1)) xx 9/color(red)(cancel(color(black)(4))) = color(red)(4)cancel(color(blue)((4x + 1))) xx 12/color(blue)(cancel(color(black)(4x + 1)))#

#9(4x + 1) = 48#

Next, Expand the terms in parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#(9 xx 4x) + (9 xx 1) = 48#

#36x + 9 = 48#

Then, subtract #color(red)(9)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#36x + 9 - color(red)(9) = 48 - color(red)(9)#

#36x + 0 = 39#

#36x = 39#

Now, divide each side of the equation by #color(red)(36)# to solve for #x# while keeping the equation balanced:

#(36x)/color(red)(36) = 39/color(red)(36)#

#(color(red)(cancel(color(black)(36)))x)/cancel(color(red)(36)) = (3 xx 13)/color(red)(3 xx 12)#

#x = (color(red)(cancel(color(black)(3))) xx 13)/color(red)(color(black)(cancel(color(red)(3))) xx 12)#

#x = 13/12#

May 10, 2017

#x=13/12#

Explanation:

Solve:

#9/4=(12)/(4x+1)#

Cross multiply the denominators.

#9(4x+1)=12xx4#

Expand.

#36x+9=12xx4#

Simplify.

#36x+9=48#

Subtract #9# from both sides.

#36x+color(red)cancel(color(black)(9))-color(red)cancel(color(black)(9))=48-9#

#36x=39#

Divide both sides by #36#.

#(color(red)cancel(color(black)(36))x)/color(red)cancel(color(black)(36))=39/36#

Simplify.

#x=39/36#

Reduce the fraction.

#3# goes into both #39# and #36#.

#x=(39-:3)/(36-:3)#

Simplify.

#x=13/12#